Ignoramus29580 wrote:
On Tue, 13 Dec 2005 14:12:39 -0800, Jim Stewart wrote:
Ignoramus29580 wrote:
On Tue, 13 Dec 2005 12:38:48 -0800, Jim Stewart wrote:
Ignoramus29580 wrote:
Let's say that I have a cable and I want to measure the AC current
going through it. Up to, say, 100 amps.
I could use a current transformer, right?
If I have a say 200:1 current transformer, then on a 100 amp AC
current it would want to produce a 0.5 amp current. Then if I stick,
say, a 1 ohm resistor across it, it would produce 0.5*1 = 0.5 volts
AC across the resistor.
Is that right?
Yes, assuming that you haven't exceeded the primary
or secondary current ratings for the transformer.
Thanks.
Furthermore, since you're probably looking to
use it on your tig welder inverter, the transformer
probably should be rated to be accurate for
pulse work rather than just 60hz sine waves.
No, I will use it for my phase converter. For measuring current on the
welder, I have a DC current meter on the welder's panel.
Here's more than you want to know about CT's:
http://www.kappaelectricals.com/technical.html#current
Oh, and don't ever leave the secondary open with
a load in the primary....
Thanks, that was all very helpful.
I have some CTs and I want to use them with some resistors and diodes
to measure several currents in my phase converter, like input current
and leg 3 current.
I would use this meter (cost $19.99 on ebay):
http://oeiwcs.omron.com/webapp/comme...6&prrfnbr=4135
My schematic would be
Diode with low voltage drop
CT pole 1 ------------||----------------------------|--
__ to Omron
Resistor Capacitor ~~ process meter
CT pole 2 -------------------------------------------|--
I need the diode because this meter is for DC readings.
Since the voltage reading is screwed up by the voltage drop of the
diode, I would need to find a diode with low voltage drop.
The problem is way worse than that. Your circuit
will measure the *peak* voltage of *1/2* of the
sine wave (less the diode drop).
I see. I coud use a 4 diode bridge and subtract 2 diode drops.
Just get a standard 5A full scale AC current meter. There's a few
problems with this diode stuff. If you want to measure upto 100A
get a 100:5 current transformer. Connect it straight to the meter.
1) they don't less than about 0.6 volts though with one diode
2) using one diode like in the ASCII art diagram will magnetize the
current transformer core.
3) the full rectifier will have a drop of 1.2 volts. You will not be
able to read any voltages below this.
A 10VA current transformer can only output upto 2 volts at 5A. SO,
with a 100:5 transformer and 50A though the wire though the donut, you
won't even get a DC voltage out of a bridge rectifier.
The reason is that the diode will only pass one
side of the sine wave. But when it's passing
that side, the capacitor will charge up to the
peak value (rms * sqt(2)) For the nitpickers,
yes, I'm assuming a well-formed sine wave.
Yes, it will be a normal sine wave.
The input impedance of the meter would also have
an effect, but since it's an unknown, I won't
go there.
It is a "known", actually, there is a table of impedances in the
datasheet. 1 or 10 megohm depending on the range.
So, 10 volts in would be 7.07 volts out.
Yes, that's OK with me. This meter can be programmed to perform linear
conversion y=ax+b.
All I have to do is bring various measurements (input current, input
voltage) to similar scale, so that:
1) when I switch to voltage (say 240V), my meter displays "240",
2) when I switch to current (say 25A), the meter displays "25".
That means that I have to perform conversion of my voltages from the
AC line voltage and from current transformers by bringing them to the
same scale. It's a simple algebra problem and can be done with
appropriate voltage dividers and also diodes.
The good news is that many times these process meters have scale and
offset values that can be programmed.
Just as mine is.
You could change the scale to correct for the 1/2 wave and cap and
change the offset to correct for the diode. Not sure it's the best
way, but if it tracked a clip-on ammeter, it may be good enough.
I agree. I will mess around with it, hopefuly it is not going to be
too difficult.
i